# Find Matrix Representation of Linear Transformation From $\R^2$ to $\R^2$

## Problem 370

Let $T: \R^2 \to \R^2$ be a linear transformation such that

\[T\left(\, \begin{bmatrix}

1 \\

1

\end{bmatrix} \,\right)=\begin{bmatrix}

4 \\

1

\end{bmatrix}, T\left(\, \begin{bmatrix}

0 \\

1

\end{bmatrix} \,\right)=\begin{bmatrix}

3 \\

2

\end{bmatrix}.\]
Then find the matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ for every $\mathbf{x}\in \R^2$, and find the rank and nullity of $T$.

(*The Ohio State University, Linear Algebra Exam Problem*)

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## Solution.

The matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ is given by

\[A=[T(\mathbf{e}_1), T(\mathbf{e}_2)],\]
where $\mathbf{e}_1=\begin{bmatrix}

1 \\

0

\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}

0 \\

1

\end{bmatrix}$ are standard basis of $\R^2$.

Since the vector $T(\mathbf{e}_2)$ is given, it remains to find $T(\mathbf{e}_1)$.

By inspection, we obtain the linear combination

\[\begin{bmatrix}

1 \\

0

\end{bmatrix}=\begin{bmatrix}

1 \\

1

\end{bmatrix}-\begin{bmatrix}

0 \\

1

\end{bmatrix}.\]
Thus, we have

\begin{align*}

T(\mathbf{e}_1)&=T\left(\,\begin{bmatrix}

1 \\

1

\end{bmatrix}-\begin{bmatrix}

0 \\

1

\end{bmatrix} \,\right)\\

&=T\left(\, \begin{bmatrix}

1 \\

1

\end{bmatrix} \,\right)-T\left(\, \begin{bmatrix}

0 \\

1

\end{bmatrix} \,\right) && \text{by linearity of $T$}\\

&=\begin{bmatrix}

4 \\

1

\end{bmatrix}-\begin{bmatrix}

3 \\

2

\end{bmatrix}\\

&=\begin{bmatrix}

1 \\

-1

\end{bmatrix}.

\end{align*}

It follows that the matrix $A$ is given by

\[A=\begin{bmatrix}

1 & 3\\

-1& 2

\end{bmatrix}.\]

The rank and nullity of $T$ are the same as the rank and nullity of $A$.

We reduced the matrix $A$ by elementary row operations as follows:

\begin{align*}

A\xrightarrow{R_2+R_1} \begin{bmatrix}

1 & 3\\

0& 5

\end{bmatrix}

\xrightarrow{\frac{1}{5}R_2}

\begin{bmatrix}

1 & 3\\

0& 1

\end{bmatrix}

\xrightarrow{R_1-3R_2}

\begin{bmatrix}

1 & 0\\

0& 1

\end{bmatrix}.

\end{align*}

Hence the rank of $A$ is $2$ (because there are two non zero rows). The nullity of $A$ is determined by the rank nullity theorem

\[\text{rank of $A$} + \text{ nullity of $A$}=2 \text{ (the number of columns of $A$)}.\]
Hence the nullity of $A$ is $0$.

In a nutshell, the rank of $T$ is $2$, and the nullity of $T$ is $0$.

## Linear Algebra Midterm Exam 2 Problems and Solutions

- True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
- Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
- Problem 2 and its solution: Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
- Problem 3 and its solution: Orthonormal basis of null space and row space
- Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
- Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
- Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$
- Problem 7 and its solution (current problem): Find matrix representation of linear transformation from $R^2$ to $R^2$
- Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space

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